If 1 kg of mass is converted to energy and used to accelerate a second 1 kg mass, what would be the final velocity?

If you take a mass $m=1:rm kg$ and turn it straight into energy, by whatever means, then indeed you will get an energy $E=mc^2$ (a.k.a. its rest energy) out of it.

If you then use all of that energy to accelerate a second $1:rm kg$ mass, that second mass will definitely be accelerated to relativistic domains. This means that you cannot use the old relationship $E_mathrm{kin}=frac12 mv^2$ between the particle's velocity $v$ relative to its initial rest frame and its kinetic energy $E_mathrm{kin}$. Instead, you need to use the relativistic version for the total energy, which reads $$ E=gamma mc^2 = frac{mc^2}{sqrt{1-v^2/c^2}}, $$ where $gamma=1/sqrt{1-v^2/c^2}$ is known as the Lorentz factor and reduces to $1$ at zero velocity (thereby giving the rest energy $E=mc^2$).

In your case, the second mass already has its rest energy, and you're doubling this, so $$ E=gamma mc^2 = 2mc^2, $$ which then gives you the equation $$frac{1}{sqrt{1-v^2/c^2}}=2$$ that you can solve for $v$ to give $v=frac{sqrt{3}}{2}capprox 0.8660,c$.